Parametric Spectral Estimation
1
ARMA Models
The most common model used in parametric spectral estimation
is the rational function model used to describe ARMA systems
q
k
b
z
k
(
)
H (z) =
=
A(z)
1+ a z
B z
k =0
p
k =1
k
k
with the corresponding difference equation
p
q
k =1
k =0
x n = ak x n k + bk w n k 2
ARMA Models
The power spectral density is
( )
( ) ( )
G xx F = G ww F H F
2
If the excitation w is a zero-mean white noise process then
( )
G xx F = 2
ww
( )
H F
2
=
( )
A (e
)
B e
2
ww
j 2 F
2
j 2 F
The most common form of model is the AR model in which
q = 0 and b0 = 1 because it usually results in the fewest parameters
needed to represent a process. In that case
( )
G xx F =
(
2
ww
A e
j 2 F
)
2
3
ARMA Models
The normal equations
p
R xx l = a p k R xx l k , l = 1,2,, p
k =1
can be use to find the coefficients a.
4
ARMA Models
Example
Let the autocorrelation of the response of a system to an
applied white noise signal with variance w2 = 6 have the
following values
R xx 0 = 4 , R xx 1 = 2 , R xx 2 = 3
Model the system producing this signal as an AR(2) system
and graph the power spectral density of the signal.
The normal equations are
R xx 0 R xx 1 a 4 1 R xx 1 =
R xx 1 R xx 0 a 4 2 R xx 2 5
ARMA Models
Example
4 2 a 4 1 2 a 4 1 0.1667 = =
3
0.667
2
4
a 4 2 a 4 2 1
1
H z =
=
A z 1 0.1667z 1 667z 2
()
()
The power spectral density is
( )
6
G xx F =
G xx
( )
1 0.1667e
j 2 F
0.667e
j 4 F
2
6
F =
1.473 0.111cos 2 F 1.333cos 4 F
(
)
(
)
6
ARMA Models
Example
7
MA Models
()
For an MA q system the transfer function is
()
()
q
H z = B z = bk z k
k =0
q
and the impulse response is h n = bk n k . So the
k =0
squared magnitude of the transfer function is
() ( )
() ( )
() H z H z 1 = B z B z 1 = D z =
q
k = q
dk z k
q
where
d n k = h n h n .
k = q
k
8
MA Models
(
)
( h n h n )
( h n h n )
h n h n n=0
= b02 + b12 + + bq2
n=1
= b0 b1 + b1b2 + + bq1bq = h n h n n=2
= b0 b2 + b1b3 + + bq2
( h n h n )
q m
n= m
=
bb
k =0
k k+m
(
)
b = ( h n h n )
q
n=1
n=2
, m q
9
MA Models
The autocorrelation of x is
R xx m = R ww m h m h m = w2 h m h m q m
R xx m = w2 bk bk + m , m q
k =0
R xx m = w2 d m
Therefore
( ) R
G xx F =
m=
j 2 Fm
e
m
=
xx q
m= q
R xx m e j 2 Fm
10